Application of Derivatives 4 Question 11
12. Let $f(x)=x^{2}+\frac{1}{x^{2}}$ and $g(x)=x-\frac{1}{x}, x \in \mathbf{R}-{-1,0,1}$. If $h(x)=\frac{f(x)}{g(x)}$, then the local minimum value of $h(x)$ is
(a) 3
(b) -3
(c) $-2 \sqrt{2}$
(d) $2 \sqrt{2}$
(2018 Main)
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Answer:
Correct Answer: 12. (c)
Solution:
- We have,
$$ \begin{aligned} & f(x)=x^{2}+\frac{1}{x^{2}} \text { and } g(x)=x-\frac{1}{x} \Rightarrow h(x)=\frac{f(x)}{g(x)} \\ & \therefore \quad h(x)=\frac{x^{2}+\frac{1}{x^{2}}}{x-\frac{1}{x}}=\frac{x-\frac{1}{x}+2}{x-\frac{1}{x}} \\ & \Rightarrow \quad h(x)=x-\frac{1}{x}+\frac{2}{x-\frac{1}{x}} \\ & x-\frac{1}{x}>0, \quad x-\frac{1}{x}+\frac{2}{x-\frac{1}{x}} \in[2 \sqrt{2}, \infty) \\ & x-\frac{1}{x}<0, \quad x-\frac{1}{x}+\frac{2}{x-\frac{1}{x}} \in(-\infty, 2 \sqrt{2}] \end{aligned} $$
$\therefore$ Local minimum value is $2 \sqrt{2}$.
